can u solve
Asked by | 5th May, 2009, 10:31: PM
From the question we gather that the required number will be 17 less than the smallest common multiple (LCM)of 520 and 468.
The L.C.M of the given numbers =product of the numbers /hcf=4680
So we need to find the number which is less than 4680 and less by 17.
So the reqd number is 4680-17=4663
If we add 17 to 4663 ,
which is exactly divisible by 520 and 468
Answered by | 7th May, 2009, 02:41: PM
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